Question

A random variable X has the following probability distribution

X	0	1	2	3	4	5	6	7
P(X)	0	k	2k	2k	3k	k2	2k2	7k2+k

then value of E(X) is:

• A. 1.66
• B. 2.66
• C. 3.66
• D. 4.66

Answer 1

The Correct Option is C

To find the expected value \( E(X) \) of the random variable \( X \), we use the formula for the expected value of a discrete random variable:

\( E(X) = \sum X_i \cdot P(X_i) \)

First, ensure that the total probability adds up to 1:

\( P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1 \)

Simplify the equation:

\( 8k^2 + 8k = 1 \)

Solve for \( k \):

\( 8k^2 + 8k - 1 = 0 \)

Using the quadratic formula \( k = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a=8 \), \( b=8 \), and \( c=-1 \):

\( k = \frac{-8 \pm \sqrt{8^2-4 \times 8 \times (-1)}}{2 \times 8} \)

\( k = \frac{-8 \pm \sqrt{64+32}}{16} \)

\( k = \frac{-8 \pm \sqrt{96}}{16} \)

\( k = \frac{-8 \pm 4\sqrt{6}}{16} \)

\( k = \frac{-2 \pm \sqrt{6}}{8} \)

Selecting the positive solution, we have:

\( k = \frac{\sqrt{6} - 2}{8} \)

Now, substitute \( k \) back into the expected value formula:

\( E(X) = 1 \cdot k + 2 \cdot 2k + 3 \cdot 2k + 4 \cdot 3k + 5 \cdot k^2 + 6 \cdot 2k^2 + 7 \cdot (7k^2 + k) \)

Substitute the calculated \( k \) value, and simplify:

\( E(X) = 1 \cdot \left(\frac{\sqrt{6} - 2}{8}\right) + 2 \cdot \left(2 \cdot \frac{\sqrt{6} - 2}{8}\right) + 3 \cdot \left(2 \cdot \frac{\sqrt{6} - 2}{8}\right) + 4 \cdot \left(3 \cdot \frac{\sqrt{6} - 2}{8}\right) + 5 \cdot \left(\left(\frac{\sqrt{6} - 2}{8}\right)^2\right) + 6 \cdot \left(2 \cdot \left(\frac{\sqrt{6} - 2}{8}\right)^2\right) + 7 \cdot \left(7\left(\frac{\sqrt{6} - 2}{8}\right)^2 + \left(\frac{\sqrt{6} - 2}{8}\right)\right) \)

By calculating the above expression, the expected value \( E(X) = 3.66 \).

Thus, the value of \( E(X) \) is: 3.66